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I have been wondering for a while that polynomials aren't just functions they are more than that. I think it's quite convincing to think of polynomials as functions adjoined with a finite sequence in some field $\mathbb{F}$. For instance, I would consider the ordered pair $(p,\{a_{n}\}_{n=0}^{N})$ to be a polynomial where p is a function and $p(x)=\sum_{n=0}^{N}a_nx^n$ for all $x$ in $\mathbb{F}$. The reason behind doing this is that I can think of distinct polynomials that represent the same function. Am I thinking in the right direction?

Anonymous
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  • Polynomials form a ring. – user14717 May 18 '20 at 19:11
  • the set of polynomials of certain maximal fixed degree over a ring or a field of coefficients form a module or a vector space, respectively, but also a commutative algebra – janmarqz May 18 '20 at 19:14
  • @DonThousand If $F$ is finite, there are finitely many functions $F\to F$ and infinitely many polynomials (the degree is unbounded) so simple counting does the job. – Mark Bennet May 18 '20 at 20:56
  • For me, polynomials have a simplicity of structure which makes them useful analytically (approximating real functions by polynomials - Taylor Series) and an algebraic richness (eg in relation to properties like field extensions and integrality) which makes them algebraically significant. You are showing a true mathematical interest in abstracting a definition which captures the core property you are interested in - of course this has been done before, and you may get some references. But it is significant to me that your mode of thought shows a mathematical instinct which is worth developing. – Mark Bennet May 18 '20 at 20:58
  • @MarkBennet Right, I should have specified characteristic 0, my bad. – Rushabh Mehta May 19 '20 at 13:19

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Yes, for example if the coefficients are from the ring $\mathbb{Z}_P$, $p$ a prime, to the two polynomials $X$ and $X^p$ is associated the same polynomial function.

Maryam
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Basically, polynomials are not functions: they're sequences of coefficients $(a_0, a_1, a_2,\dots, a_n,\dots)$ from a commutative ring $R$, with finite support, i.e. such that only a finite number is nonzero. The set of such sequences is denoted $R^{(\mathbf N)}$ and is endowed with a componentwise addition, and scalar multiplication, which makes it an $R$-module. Furthermore, a product $(a_n)_{n\in\mathbf N}\cdot(b_n)_{n\in\mathbf N}=(c_n)_{n\in\mathbf N}$ is defined, where $$\forall n\in\mathbf N,\quad c_n=\sum_{i+j=n}a_ib_j.$$ It happens that, if we denote $X$ the particular sequence $(0,1,0,\dots,0,\dots)$, each polynomial $P=(a_0, a_1, \dots, a_d, 0,\dots,0,\dots)$, where $a_d $ is the last nonzero coefficient, we can write $$P=a_0+a_1X+\dots+a_d X^d,$$ in a unique way.

To this polynomial is associated a polynomial function \begin{align} p:R&\longrightarrow R \\ r&\longmapsto p(r)=a_0+a_1r+\dots+a_dr^d, \end{align} and this correspondence is bijective if the ring $R$ has characteristic $0$.

J. W. Tanner
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Bernard
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    @J.W. Tanner; Thanks for editing my answer. I should definitely check more carefully my typos! – Bernard May 18 '20 at 22:51
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In an infinite field there exists a bijection between the polynomial and polynomial function so one can regard polynomials as functions but unfortunately that doesn't happens in arbitrary fields (Friedberg Linear Algebra 4th editions explains that in page 569). Consider for instance $Z_2$.

Anonymous
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